The value of $\sqrt{8}$ lies between which two consecutive integers ? Integers that appear in order when counting, for example 2 and 3.
Explanation: Consider the perfect squares near $8$ . [ What are perfect squares? Perfect squares are integers which can be obtained by squaring an integer. The first 13 perfect squares are: $ 1,4,9,16,25,36,49,64,81,100,121,144,169$ $4$ is the nearest perfect square less than $8$ $9$ is the nearest perfect square more than $8$ So, we know $4 < 8 < 9$ So, $\sqrt{4} < \sqrt{8} < \sqrt{9}$ So $\sqrt{8}$ is between $2$ and $3$.